Is there any relationship between Cauchy-Riemann equations and vector fields on manifolds? Well, suppose we have $f : \mathbb{C} \to \mathbb{C}$ analytic, then if $f = u + iv$ the functions $u,v : \mathbb{C} \to \mathbb{R}$ satisfy the Cauchy-Riemann equations: $D_1u=D_2v$ and $D_2u=-D_1v$. Now, suppose we pick $(x,\mathbb{C})$ the cartesian coordinates in $\mathbb{C}$. Then we have:
$$\begin{cases}\dfrac{\partial u}{\partial x^1}&=\phantom{-}\dfrac{\partial v}{\partial x^2} \\ \\ \dfrac{\partial u}{\partial x^2} &= -\dfrac{\partial v}{\partial x^1}\end{cases}$$
This can also be written as simply:
$$i\dfrac{\partial f}{\partial x^1}=\dfrac{\partial f}{\partial x^2}$$
But now here is the interesting thing I've noticed. When we work with arbitrary smooth manifolds, the partials operators relative to some coordinate system are tangent vectors to the coordinate lines. So that $\partial /\partial x ^1$ and $\partial/\partial x^2$ are tangent vectors to the coordinate lines. When we work with points of $\mathbb{C}$ and we understand then as vectors (identifying the tangent space at the origin with the space $\mathbb{C}$ itself), multiplying by $i$ is the same as rotating a vector by $\pi/2$.
In the standard cartesian coordinates, $\partial/\partial x^1$ is pointing in the direction of the $x$ axis and $\partial/\partial x^2$ is pointing in the direction of the $y$ axis. In that case, $\partial /\partial x^2$ is simply $\partial /\partial x^1$ rotated $\pi/2$ in the counterclockwise direction. And using $i$ to express rotations, this is exactly what is written up there.
I'm not sure if I've made myself clear, but the question is: "is there any relationship between the Cauchy-Riemann equations and the vector fields defined in $\mathbb{C}$ as a smooth manifold that gives us deeper understanding of what analytic functions do when transforming $\mathbb{C}$ into another $\mathbb{C}$?"
Thanks very much in advance!
 A: Let $f(x,y)=u(x,y)+iv(x,y)$ be an analytic function. The vector field 
$$(x,y)\mapsto V(x,y)=(u(x,y),v(x,y))$$
is then both irrotational (i.e. $\operatorname{curl}(V)=0$) and solenoidal ($\operatorname{div}(V)=0$). These are the "physical" implications at the level of vector fields one can deduce from the the C.R. equations. 
On viceversa: in classical potential theory, one usually studies complex valued vector fields $(x,y)\mapsto V(x,y)=\left(u(x,y),v(x,y)\right)$ by considering the linear combination
$$f(x,y)=u(x,y)+iv(x,y),$$
on the whole complex plane. If the vector field $V(x,y)$ is both irrotational and solenoidal, then $f$ satisfies the C.R. equations (which does not imply analyticity).


*

*Small remark


Let $z=x+iy$ and $f$ as above. 
Using the C.R. equations on $u$ and $v$ one gets the following chain of equivalences
$$ 
u ~\text{and}~ v ~\text{satisfy the C.R. equations}\Leftrightarrow \omega:=f(z)dz~\text{is closed}\Leftrightarrow \tilde{V}=(f,if)~\text{is irrotational}.
$$
